9-7: FACTORING SPECIAL CASESEssential Question: Name and give examples of the two types of special cases you learned to factor

9-7: Factoring Special Cases

Back in section 9-4, we found the square of a binomial (a + b)2 = (a + b)(a + b) = a2 + 2ab + b2 and

(a – b)2 = (a – b)(a – b) = a2 – 2ab + b2

Any trinomial of the form a2 + 2ab + b2 or a2 – 2ab + b2 is a perfect square trinomial. You can factor a perfect square trinomial into identical binomial factors. Examples:

x2 + 10x + 25 = (x + 5)(x + 5) = (x + 5)2

x2 – 10x + 25 = (x – 5)(x – 5) = (x – 5)2

9-7: Factoring Special Cases

You can factor perfect square trinomials like we did two weeks ago. Or you can recognize a perfect-square trinomial and factor it quickly. Here are the two things to look for to recognize a perfect-square trinomial. The first and last terms can be written as

the product of two identical terms The middle term is twice the product of

each of the squared factors

9-7: Factoring Special Cases

Consider the following two trinomials: 4x2 + 12x + 9 4x2 + 20x + 9

4x2 = 2x ● 2x 4x2 = 2x ● 2x 9 = 3 ● 3 9 = 3 ● 3 2(2x ● 3) = 12x 2(2x ● 3) ≠ 20x

Perfect Square TrinomialNot a Perfect Square Trinomial

(2x + 3)(2x + 3) or (2x + 3)2

When you factor a perfect square trinomial, it may help to write the first and last terms

9-7: Factoring Special Cases

Example 1: Factoring a Perfect Square Trinomial with a = 1 Factor x2 – 8x + 16

x2 – 8x + 16 = x ● x – 8x + 4 ● 4 Check the middle term 2(x ● 4) = 8x It is a perfect square trinomial

(x – 4)2

YOUR TURN: Factor each expression x2 + 8x + 16 n2 + 16n + 64

(x + 4)2

(n + 8)2

9-7: Factoring Special Cases

Example 2: Factoring a Perfect Square Trinomial with a ≠ 1 Factor 9g2 + 12g + 4

9g2 + 12g + 4 = 3g ● 3g + 12g + 2 ● 2 Check the middle term 2(3g ● 2) = 12g It is a perfect square trinomial

(3g + 2)2

YOUR TURN: Factor each expression 9d2 – 12d + 4 4t2 + 36t + 81

(3d – 2)2

(2t + 9)2

9-7: Factoring Special Cases

Back in section 9-4, we found the difference of two squares (a + b)(a – b) = a2 – b2

In a difference of perfect square problem, there is no middle term, so you’re looking to find two numbers that add to 0. If you decide to shortcut, the two things to look for to recognize a difference of perfect squares are: The first and last terms can be written as the

product of two identical terms There is no middle term

9-7: Factoring Special Cases

Example 3: The difference of two squares for a = 1 Factor x2 – 64

x2 – 64 = x ● x – 8 ● 8 (x + 8)(x – 8)

YOUR TURN: Factor each expression x2 – 36 m2 – 100

(x + 6)(x – 6)(m + 10)(m – 10)

9-7: Factoring Special Cases

Example 4: The difference of two squares for a ≠ 1 Factor 4x2 – 121

4x2 – 121 = 2x ● 2x – 11 ● 11 (2x + 11)(2x – 11)

YOUR TURN: Factor each expression 9v2 – 4 25w2 – 64

(3v + 2)(3v – 2)(5w + 8)(5w – 8)

9-7: Factoring Special Cases

Example 5: Factoring out a Common Factor Sometimes, polynomials may not look like

the follow one of these patters, but may appear if a GCF is factored out first

Factor 10x2 – 40 10x2 – 40 = 10(x2 – 4)

= 10(x + 2)(x – 2) YOUR TURN: Factor each expression

8y2 – 50 3c2 – 75

2(2y + 5)(2y – 5)3(c + 5)(c – 5)

9-7: Factoring Special Cases

Assignment Worksheet #9-7 Problems 1 – 33, odds